我有一个有帐号和卡号的数据库。我将这些匹配到一个文件,以将任何卡号更新为帐号,这样我只使用帐号。

我创建了一个将表链接到帐户/卡数据库的视图,以返回table ID和相关的帐号,现在我需要更新那些ID与account number匹配的记录。

这是Sales_Import表,其中的帐号字段需要更新:

LeadID AccountNumber
147 5807811235
150 5807811326
185 7006100100007267039

这是RetrieveAccountNumber表,我需要从这里更新:

LeadID AccountNumber
147 7006100100007266957
150 7006100100007267039

我尝试了下面的方法,但到目前为止运气都不佳:

UPDATE [Sales_Lead].[dbo].[Sales_Import] 
SET    [AccountNumber] = (SELECT RetrieveAccountNumber.AccountNumber 
                          FROM   RetrieveAccountNumber 
                          WHERE  [Sales_Lead].[dbo].[Sales_Import]. LeadID = 
                                                RetrieveAccountNumber.LeadID) 

它将卡号更新为帐号,但是帐号被NULL替换


当前回答

以防表在不同的数据库中。(该)

update database1..Ciudad
set CiudadDistrito=c2.CiudadDistrito

FROM database1..Ciudad c1
 inner join 
  database2..Ciudad c2 on c2.CiudadID=c1.CiudadID

其他回答

我相信一个带JOIN的UPDATE FROM会有帮助:

MS SQL

UPDATE
    Sales_Import
SET
    Sales_Import.AccountNumber = RAN.AccountNumber
FROM
    Sales_Import SI
INNER JOIN
    RetrieveAccountNumber RAN
ON 
    SI.LeadID = RAN.LeadID;

MySQL和MariaDB

UPDATE
    Sales_Import SI,
    RetrieveAccountNumber RAN
SET
    SI.AccountNumber = RAN.AccountNumber
WHERE
    SI.LeadID = RAN.LeadID;

谢谢你的回复。但我找到了一个解决办法。

UPDATE Sales_Import 
SET    AccountNumber = (SELECT RetrieveAccountNumber.AccountNumber 
                          FROM   RetrieveAccountNumber 
                          WHERE  Sales_Import.leadid =RetrieveAccountNumber.LeadID) 
WHERE Sales_Import.leadid = (SELECT  RetrieveAccountNumber.LeadID 
                             FROM   RetrieveAccountNumber 
                             WHERE  Sales_Import.leadid = RetrieveAccountNumber.LeadID)  

将内容从一个表复制到另一个表的简单方法如下:

UPDATE table2 
SET table2.col1 = table1.col1, 
table2.col2 = table1.col2,
...
FROM table1, table2 
WHERE table1.memberid = table2.memberid

还可以添加条件以复制特定的数据。

试试这个:

UPDATE
    Table_A
SET
    Table_A.AccountNumber = Table_B.AccountNumber ,
FROM
    dbo.Sales_Import AS Table_A
    INNER JOIN dbo.RetrieveAccountNumber AS Table_B
        ON Table_A.LeadID = Table_B.LeadID 
WHERE
    Table_A.LeadID = Table_B.LeadID

我对foo也有同样的问题。对于在bar中没有匹配键的foo行,New被设置为null。我在Oracle做了类似的事情:

update foo
set    foo.new = (select bar.new
                  from bar 
                  where foo.key = bar.key)
where exists (select 1
              from bar
              where foo.key = bar.key)